Question

Evaluate
${\int }_1^2\frac{(1+\log x{)}^2}{x}dx$

Answer 1

${\int }_1^2\frac{(1+logx{)}^2}{x}$ dx put log x = t when x= 1 $\Rightarrow$ t = 0

Differntiating $\frac{1}{2}dx=dtx=2\Rightarrow t=log2$

${\int }_0^{log2}(1+t{)}^{2}dt$

$=[\frac{(1+t{)}^3}{3}{]}^{}log2$

$=\frac{(1+log2{)}^3}{3}-\frac{1}{3}=\frac{1+2log2+(log2{)}^2-1}{3}$

$=\frac{2log2+(log2{)}^2}{3}$

$=\left(log2\right)\frac{[2+log2]}{3}$